How do use the first derivative test to determine the local extrema #f(x) = (x+1)(x-3)^2#?

Answer 1

Have a look:

You can use the Product and Chain Rule to find the derivative: #f'(x)=1*(x-3)^2+(x+1)*2(x-3)=x^2-6x+9+2(x^2-3x+x-3)=# #=x^2-6x+9+2(x^2-2x-3)=x^2-6x+9+2x^2-4x-6=# #=3x^2-10x+3# Now you can set it equal to zero to find the x coordinate(s) of the local extrema: #3x^2-10x+3=0# We can use the Quadratic Formula: #x_(1,2)=(10+-sqrt(100-36))/6=(10+-8)/6=# So you get: #x_1=3# giving #y=0# #x_2=1/3# giving #y=9.5#

Graphically: graph{(x+1)((x-3)^2) [-10, 10, -5, 5]}

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Answer 2
To use the first derivative test to determine the local extrema of \( f(x) = (x+1)(x-3)^2 \), follow these steps: 1. Find the first derivative of the function, \( f'(x) \). 2. Identify critical points by setting \( f'(x) = 0 \) and solving for \( x \). 3. Determine the sign of \( f'(x) \) in intervals around the critical points. 4. Use the sign of \( f'(x) \) to determine whether the critical points correspond to local maxima, local minima, or neither. Now, let's proceed with the steps: 1. First derivative of \( f(x) \): \[ f'(x) = (x-3)^2 + (x+1) \cdot 2(x-3) = (x-3)(2x-6) + 2(x+1) \] \[ f'(x) = 2x^2 - 6x - 6x + 18 + 2x + 2 \] \[ f'(x) = 2x^2 - 10x + 20 \] 2. Find critical points by setting \( f'(x) = 0 \): \[ 2x^2 - 10x + 20 = 0 \] \[ x^2 - 5x + 10 = 0 \] This quadratic equation has no real roots, hence no critical points. 3. Determine the sign of \( f'(x) \) in intervals: - For \( x < 0 \), pick \( x = -1 \): \( f'(-1) = 2(-1)^2 - 10(-1) + 20 = 2 + 10 + 20 > 0 \) - For \( 0 < x < 3 \), pick \( x = 1 \): \( f'(1) = 2(1)^2 - 10(1) + 20 = 2 - 10 + 20 > 0 \) - For \( x > 3 \), pick \( x = 4 \): \( f'(4) = 2(4)^2 - 10(4) + 20 = 32 - 40 + 20 < 0 \) 4. Interpretation: - Since \( f'(x) > 0 \) for \( x < 0 \) and \( 0 < x < 3 \), and \( f'(x) < 0 \) for \( x > 3 \), there are no local extrema for the function \( f(x) = (x+1)(x-3)^2 \).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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